Note that $a>0$, thus I'm not sure if we can apply residues here. (For $a=0$ the integral doesn't converge).

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}$$

Despite the simple expression under the integral, I didn't find it in G-R book.

I attempted the most straightforward way - geometric series - since $e^x \geq 1$.

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty (-1)^{k+1} \int_0^\infty \frac{e^{-kx}dx}{a+x}$$

The integrals on the right can be expressed using incomplete gamma function:

$$\int_0^\infty \frac{e^{-kx}dx}{a+x}=\frac{e^{ka}}{k} \Gamma(0,ka)$$

We obtain:

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty (-1)^{k+1} \frac{e^{ka}}{k} \Gamma(0,ka)$$

This is correct, but not very useful, since each incomplete gamma just disguises an integral. It can be computed independently by:

$$\Gamma (0,t)=\cfrac{\exp(-t)}{t+1-\cfrac{1}{t+3-\cfrac{4}{t+5-\cfrac{9}{t+7-\cdots}}}}$$

Thus I suppose we can express the integral as:

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \cfrac{1}{ka+1-\cfrac{1}{ka+3-\cfrac{4}{ka+5-\cfrac{9}{ka+7-\cdots}}}}$$

Amusing, but again, not very useful.

Does this integral have a closed form? If so, what is it and how do we find it?


Another interesting expression for incomplete gamma from DLMF:

$$\Gamma (0,t)=e^{-t} \sum_{n=0}^\infty \frac{L_n (t)}{n+1}$$

Here $L_n (t)$ are Laguerre polynomials. This series converges extremely slowly (and oscillates), so it seem quite useless for computation, however gives a pretty expression for the integral:

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \sum_{n=0}^\infty \frac{L_n (ka)}{n+1} $$

Note, that the continued fraction above converges much faster than the series.

  • $\begingroup$ In the cause of a proof, I found a solution of the integral: math.stackexchange.com/questions/3433267/… $\endgroup$ – stocha Nov 23 '19 at 16:52
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    $\begingroup$ @stocha, very interesting, thank you $\endgroup$ – Yuriy S Nov 23 '19 at 18:53
  • $\begingroup$ Thanks, your contribution helped me with the proof! $\endgroup$ – stocha Nov 23 '19 at 22:18

Since: $$\mathcal{L}\left(\frac{1}{1+e^x}\right) = \frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right],\qquad \mathcal{L}^{-1}\left(\frac{1}{a+x}\right)=e^{-as}$$ the original integral equals: $$ \int_{0}^{+\infty}\frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds=\frac{1}{2}\int_{0}^{+\infty}\int_{\frac{s+1}{2}}^{\frac{s+2}{2}}\sum_{n\geq 0}\frac{e^{-as}}{(n+u)^2}\,du\,ds $$ and by integration by parts, the LHS just depends on: $$ \int_{0}^{+\infty}\left[\log\Gamma\left(\frac{s+2}{2}\right)-\log\Gamma\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds$$ that can be further expanded by exploiting the Weierstrass product for the $\Gamma$ function, leading to a fast-convergent series, whose first term is $-\frac{\gamma}{2a}$.

  • $\begingroup$ What is $\psi$? $\endgroup$ – GFauxPas Jun 10 '16 at 14:16
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    $\begingroup$ @GFauxPas, digamma function, I presume $\endgroup$ – Yuriy S Jun 10 '16 at 14:16
  • $\begingroup$ Do I understand correctly that Laplace transform in this case means just using $$\frac{1}{a+x}=\int_0^\infty e^{-(a+x)t}dt$$, reversing order of integration and then solving $$\int_0^\infty \frac{e^{-tx} dx}{1+e^x}$$? $\endgroup$ – Yuriy S Jun 10 '16 at 14:30
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    $\begingroup$ @YuriyS: absolutely right. $\endgroup$ – Jack D'Aurizio Jun 10 '16 at 14:31

Just some further thoughts, which neatly connect different methods of transforming the integral.

Using Jack D'Aurizio's answer, we work with the integral:

$$I(a)=\int_{0}^{+\infty}\frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds= \\ = \sum_{n=0}^\infty (-1)^n \int_{0}^{+\infty} \frac{e^{-a s} ds}{n+s+1}$$

This is actually the same series we get by expanding the original integrand in geometric series:

$$I(a)=\sum_{n=0}^\infty (-1)^n \int_0^\infty \frac{e^{-(n+1)x}dx}{x+a}$$

However, the first form of the series might be better for acceleration.

Using Euler's transform:

$$\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{n-k+s+1}=(-1)^n \mathrm B (n+1,s+1)$$

We have:

$$\sum_{n=0}^\infty \frac{(-1)^n}{n+s+1}=\sum_{n=0}^\infty \frac{\mathrm B (n+1,s+1)}{2^{n+1}}$$

Which gives us:

$$I(a)=\sum_{n=0}^\infty \frac{1}{2^{n+1}}\int_{0}^{+\infty} \mathrm B (n+1,s+1)e^{-a s} ds \tag{1}$$

Using the integral definition for the Beta function, we have:

$$I(a)=\sum_{n=0}^\infty \frac{1}{2^{n+1}}\int_{0}^{+\infty} \int_0^1 t^n (1-t)^s e^{-a s} dt ds $$

Summation gives us:

$$I(a)= \int_{0}^{+\infty} \int_0^1 \frac{(1-t)^s}{2-t} e^{-a s} dt ds $$

Integrating w.r.t. $s$:

$$I(a)= \int_0^1 \frac{dt}{(2-t)(a-\log(1-t))} = \int_0^1 \frac{dt}{(1+t)(a-\log t)}$$

Which, funny enough, brings us back to the original integral.

Let's get back to (1). The Beta functions and the exponential are both very nice functions, which fall to $0$ fast.

$$I(a)=\sum_{n=0}^\infty \frac{1}{2^{n+1}}\int_{0}^{+\infty} \mathrm B (n+1,s+1)e^{-a s} ds$$

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Finding a suitable approximation for the Beta functions (conventional asymptotics are nor very useful in this case), we can get a fast converging series for $I(a)$. There's probably a good approximation as a sum of two or three exponential terms.

The integration by parts advice from Jack is sound as well, but I haven't worked out the resulting series yet.

$$I(a)=\frac{\log \pi}{2} + a \int_0^\infty \left[\log\Gamma\left(\frac{s+2}{2}\right)-\log\Gamma\left(\frac{s+1}{2}\right)\right]e^{-as} ds$$


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